As a result of friction, the angular speed of a wheel changes with time according to

d(theta)/dt = w0 e^-(aplpha*t)
where w0 and alpha are constants. The angular speed changes from an initial angular speed of 4.87 rad/s to 4.27 rad/s in 33.93s.
A) Determine the magnitude of the angular acceleration after 2.46s. answer in rad/s^2
B) How many revolutions does the wheel make after 2.59 s? answer in rev
C) find the number of revolutions it makes before coming to rest, answer in rev.

any help would be appreciated, thank you in advance

To solve this problem, we will use the given equation

(dθ/dt) = w0 * e^(-αt)

We are given the following information:
Initial angular speed, ω0 = 4.87 rad/s
Final angular speed, ω = 4.27 rad/s
Time, t = 33.93 s

A) Determine the magnitude of the angular acceleration after 2.46 s.

To find the angular acceleration, we need to differentiate the given equation with respect to time:

d^2(θ)/dt^2 = d/dt (w0 * e^(-αt))
= -α * w0 * e^(-αt)

Now, substitute the values:
t = 2.46 s
w0 = 4.87 rad/s
α = unknown

We are given the initial angular speed, so we can use it to find α:

4.87 = w0 * e^(-αt)

Solving for α:

e^(-αt) = 4.87 / w0
αt = ln(4.87 / w0)
α = ln(4.87 / w0) / t

Substituting the values:
α = ln(4.87 / 4.87) / 2.46
α ≈ 0 rad/s^2 (since ln(1) = 0)

Therefore, the magnitude of the angular acceleration after 2.46 s is approximately 0 rad/s^2.

B) How many revolutions does the wheel make after 2.59 s?

To find the number of revolutions, we need to integrate the given equation with respect to time:

Δθ = ∫(w0 * e^(-αt)) dt
= w0 * ∫e^(-αt) dt

Integrating and substituting the values:
t = 0 to t = 2.59 s

Δθ = w0 * ∫e^(-αt) dt
= w0 * [-1/α * e^(-αt)]
= -w0/α * (e^(-α*2.59) - e^(-α*0))

Since we found that α ≈ 0, the equation becomes:

Δθ = -w0/α * (e^(0) - e^(0))
= -w0/α * (1 - 1)
= 0

Therefore, the wheel does not make any revolutions after 2.59 s.

C) Find the number of revolutions it makes before coming to rest.

To find the number of revolutions the wheel makes before coming to rest, we need to find the time when the angular speed becomes zero.

Given: d(θ)/dt = w0 * e^(-αt)

At angular speed ω = 0:
0 = w0 * e^(-αt)

Solving for t:
e^(-αt) = 0 / w0
t = ln(0 / w0)
t = -∞

The angular speed becomes zero at time t = -∞, which means the wheel does not come to rest within the time frame provided. Therefore, it does not make any complete revolutions before coming to rest.

To solve this problem, we need to use the given equation for the angular speed with respect to time and solve for the different parts.

Given equation: d(theta)/dt = w0 e^-(alpha*t)

A) To determine the magnitude of the angular acceleration after 2.46s, we need to find the second derivative of the angular displacement with respect to time.

Taking the derivative of the given equation, we get:

d^2(theta)/dt^2 = -w0*alpha*e^-(alpha*t)

To calculate the angular acceleration at t = 2.46s, we substitute t = 2.46 into the equation:

d^2(theta)/dt^2 = -w0*alpha*e^-(alpha*2.46)

B) To find the number of revolutions the wheel makes after 2.59s, we first determine the change in angular displacement between 2.59s and the initial time.

Δθ = ∫(w0 e^-(alpha*t)) dt
= ∫(w0 e^-(alpha*t)) dt from t = 0 to t = 2.59

We integrate the expression with respect to time from 0 to 2.59. This will give us the change in angular displacement.

C) To find the number of revolutions the wheel makes before coming to rest, we need to solve for the time at which the angular speed becomes zero and then find the total angular displacement until that time.

Given equation: d(theta)/dt = w0 e^-(alpha*t)

Set the angular speed to zero: 0 = w0 e^-(alpha*t)

Solve for t by taking the natural logarithm:

ln(0) = ln(w0 e^-(alpha*t))
0 = ln(w0) - alpha*t

Solving for t gives: t = ln(w0)/alpha

Now we can find the total angular displacement from t = 0 to t = ln(w0)/alpha:

Δθ = ∫(w0 e^-(alpha*t)) dt
= ∫(w0 e^-(alpha*t)) dt from t = 0 to t = ln(w0)/alpha

We integrate the expression with respect to time from 0 to ln(w0)/alpha. This will give us the total angular displacement before the wheel comes to rest.

angular speed=dTheta/dt

angular acceleration=d"theta/dt"
= -alpha*wo e^{-alpha*t)

now given: speed(t=0)=wo=4.87
speed(t=33.93)=4.87e^(-alpha*33.93) solve for alpha
take the ln of each side,
ln (4.27)=ln(4.87)-alpha*33.93
solve for alpha

for b, integrate dTheta/dt from t=0 to t=2.57

for c. rest means dTheta/dt is zero, so solve for t when dTheta/dt=0, notice that it NEVER gets to rest, so tf=infinity.